Question

sin theta = cos theta/2 prove when 0°< theta< 360°​

Answer 1

Answer:

$sin \alpha = \cos( \frac{ \alpha }{2} )$

$\cos(90 - \alpha ) = \cos( \frac{ \alpha }{2} )$

$90 - \alpha = \frac{ \alpha }{2}$

$\frac{3 \alpha }{2} = 90$

$\alpha = 60$